 Solution

Write the Schwarzian derivative asThe first prolonged generator is ${\partial }_x$, and $S$ has no explicit dependence on $x$, so ${\mathrm{pr}}^{(3)}{V}_1(S)=0$. Under ${\mathrm{pr}}^{(3)}{V}_2$, each $u_j$ has weight $-j$. Both terms in $S$ therefore have weight $-2$, giving ${\mathrm{pr}}^{(3)}{V}_2(S)=-2S$. Substitution offor $V_3$ gives ${\mathrm{pr}}^{(3)}{V}_3(S)=-4xS$. Therefore the projective transformation weights of the Schwarzian derivative are

For an equation $\Delta =S-f(x)=0$, the Lie point symmetry of an ordinary differential equation criterion is